Mathcounts Chapter Competition Solutions Worksheet - Middle School - 2014 Page 6
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11
12represent the side lengths of a triangle?
the numbers in 424242…42 is divisible by 3,
The sum of the lengths of two sides of a
so adding or subtracting 1 results in a number
triangle must be greater than the other side.
that is not divisible by 3.
Start with a smallest side of 2:
5: For a number to be divisible by 5, it must
(2, 3, 4) (2, 6, 7)
end in 0 or 5. As previously stated, a number
(2, 4, 5) (2, 7, 8)
of this form will have a ones digit of 3 or 1,
(2, 5, 6) -- 5 values
and is therefore not divisible by 5.
Start with a smallest side of 3:
7: As with 3, the number 424242…42 is
(3, 4, 5) (3, 5, 6) (3, 6, 7) (3, 7, 8)
divisible by 7. Therefore, adding or
(3, 4, 6) (3, 5, 7) (3, 6, 8) -- 7 values
subtracting 1 will result in a number that is not
Start with a smallest side of 4:
divisible by 7.
(4, 5, 6) (4, 6, 7) (4, 7, 8)
11: We can’t immediately rule out 11 as we
(4, 5, 7) (4, 6, 8)
did the other primes, so let’s divide and see if
(4, 5, 8) -- 6 values
it is a factor of some number of the given
Start with a smallest side of 5:
form. Dividing, we find that
(5, 6, 7) (5, 7, 8)
424242424243/11 = 38567493113 and
(5, 6, 8) -- 3 values
4242424241/11 = 385674931. The
Start with a smallest side of 6:
smallest prime that divides a number of this
(6, 7, 8) -- 1 value
form is11 Ans.
5 + 7 + 6 + 3 + 1 = 22 Ans.
29. Point E lies within rectangle ABCD.
27. If 1/x + 3/y = 3/4 and 3/x − 2/y = 5/12, what
IF AE = 7, BE = 5 and CE = 8, what is DE?
is x + y?
To solve this system of equations, we start by
multiplying the first equation by 2 and the
second equation by 3 to get 2/x + 6/y = 6/4
and 9/x − 6/y = 15/12.
Next, we add the two equations and get
2/x + 9/x + 6/y − 6/y = 6/4 + 15/12 →
Draw horizontal segment JK through point E
11/x = 33/12 → 33x = 132 → x = 4.
and parallel to sides AB and DC. Then draw
Substituting for x in the first equation and
vertical segment GH through point E and
solving for y yields 1/4 + 3/y = 3/4 → 3/y =
parallel to sides AD and BC. As shown, we’ll
1/2 → y = 6. So, x + y = 4 + 6 = 10 Ans.
let EG = x, EH = y, EK = w and EJ = z. It
follows that JK = z + w and GH = x + y.
Segments GH and JK are perpendicular to
28. What is the smallest prime number that
the sides of the rectangle and several right
divides some number of the form
triangles are formed. Applying the
424242…42 + 1 or 424242…42 – 1?
Pythagorean Theorem, we have :
Let’s start checking from the smallest prime
→ x
2
+ z
2
2
2
+ z
2
x
= 7
= 49
number.
→ w
w
2
+ x
2
= 5
2
2
+ x
2
= 25
2: Any number of this form will have a ones
→ w
w
2
+ y
2
= 8
2
2
+ y
2
= 64
digit that is odd, namely 3 or 1, and therefore
→ DE = √(y
2
= (DE)
2
2
2
2
y
+ z
+ z
)
will not be divisible by 2.
We can manipulate the first three equations to
3: For a number to be divisible by 3, the sum
get a new equation relating y and z.
of its digits must be divisible by 3. The sum of
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